# integration part 2

• Feb 25th 2008, 02:45 PM
jliddell
integration part 2
$
\int (xsinx^2cosx^2),dx

$

all i need are steps leading to the answer
• Feb 25th 2008, 02:46 PM
Krizalid
Start by makin' $u=x^2.$
• Feb 25th 2008, 02:58 PM
jliddell
integration part 2
ok not quite getting it. can you help me out a little more please.
• Feb 25th 2008, 03:00 PM
Krizalid
$u=x^2\implies du=2x\,dx,$

$\int {x\sin x^2 \cos x^2 \,dx} = \frac{1}
{2}\int {\sin u\cos u\,du} = \frac{1}
{4}\int {\sin 2u\,du} .$
• Feb 26th 2008, 11:57 AM
simsima_1
needing a little more help w/ this....
ok so the answer is

$

(x)tan^{-1}(x)-\frac{1}{2}ln({x^2}+1)+C
$

and i'm not sure how i would get that using

$
{1/4} \int sin2u

$
• Feb 26th 2008, 12:20 PM
Jhevon
Quote:

Originally Posted by simsima_1
ok so the answer is

$

(x)tan^{-1}(x)-\frac{1}{2}ln({x^2}+1)+C
$

and i'm not sure how i would get that using

$
{1/4} \int sin2u

$

there are often many different forms for the answer to an integral. following Krizalid's suggestion is good enough.